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Momentum and Conservation of Momentum Problem

  1. Dec 1, 2006 #1
    First of all I would like to explain that this is the last question I need to answer to complete an assignment that consisted of 20 University of Waterloo SIN Contest questions. This question is from the 1999 SIN contest. I would ask my Physics teacher for help but it is now the weekend and the assignment is due on Monday so I have come here looking for aid.

    PS. The question itself is kind of a Canadian politics joke.

    1. The problem statement, all variables and given/known data

    Big Jean Chretien. in a liberal mood. throws a carton of pepper spray at Lucien Bouchard. (Assault by pepper?). Lucien catches the carton, and hangs onto it, of course. (Has he ever returned a federal gift?). Since both men were initially standing at rest on a frictionless horizontal ice surface, this exchange causes them to separate. Each man has a mass of 100 kg, the carton has a mass of 20 kg, and the carton left Jean's hands with a horizontal velocity of 10 m/s relative to Jean's body. Calculate the final relative separation speed between the two men. (Answer in m/s).

    (a) 2.7

    (b) 3.1

    (d) 7.1

    (e) 8.9

    (c) 5.2

    2. Relevant equations

    Pt = Pt'
    Pt' = P1' + P2' (Similar to Pt)
    P1 = m1v1 (Similar to P2, P1', P2' etc...)
    m1v1 + m2v2 = m1v1' + m2v2'

    P = Momentum
    m = Mass
    v = Velocity

    3. The attempt at a solution

    Ok so, the first thing i did is find the speed of Lucien Bouchard. Since the 20kg projectile sticks to him once he catches it then we know that v1' = v2' = v'

    Knowing that:

    Pt = Pt'
    m1v1 + m2v2 = m1v1' + m2v2'
    m1v1 + m2v2 = m1v' + m2v'
    m1v1 + m2v2 = v'(m1 + m2)
    (20)(10) + (100)(0) = v'(20 + 100)
    v' = 1.67 m/s

    Now to find the speed of Jean Chretien. To find this one I used a similar equation but since he is not moving before he throws the projectile Pt = 0.

    Pt = Pt'
    0 = Pt'
    0 = m1v1' + m2v2'
    -m1v1' = m2v2'
    -(100)v1' = (20)(10)
    v1' = -2 m/s

    This would give me a final relative separation speed of:

    v = v' - (v1')
    v = (1.67) - (-2)
    v = 3.67 m/s

    And since this is not one of the multiple choice answers it means that I did something wrong somewhere.

    Any help would be greatly appreciated, especially before monday :smile:. And if there is anything you guys don't understand then please ask and I will explain more thoroughly.

    Attached is a .pdf file with a scan of the question and my rough work.

    Thanks in advance.

    - Stahs
     

    Attached Files:

  2. jcsd
  3. Dec 1, 2006 #2

    OlderDan

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    I think you know how to do the problem, but you missed a key phrase highlighted above. You used the wrong velocity for the carton.
     
  4. Dec 1, 2006 #3
    Ohhh I see, so wouldn't that make the carton's velocity 8m/s since Jean moves at 2m/s? That still doesnt give the right answer.
     
  5. Dec 1, 2006 #4

    OlderDan

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    Start at the beginning when the carton leaves Jean. If the carton is not moving at 10m/s, Jean is not moving at 2m/s. You need to redo the whole event of the carton being thrown.
     
  6. Dec 2, 2006 #5
    Hmm, I don't really know how to do it then. I can't think of any way of getting Jean's speed. Can I maybe get a hint?
     
  7. Dec 2, 2006 #6

    OlderDan

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    Jean and the carton are a system that has zero momentum. You knew that and used it to find Jean's velocity when you thought the carton's velocity was 10m/s relative to the ice. Do the same calculation, but use the fact that the carton's velocity is 10m/s relative to Jean. If the carton has velocity v relative to the ice, what is Jean's velocity relative to the ice?
     
  8. Dec 2, 2006 #7
    Ok so... Jean's velocity relative to the box is still 2 m/s right? Which means that his velocity relative to the ice is less than 2 m/s and that the box's velocity relative to the ice is 8 m/s?
     
  9. Dec 2, 2006 #8
    But i still have no idea how to get his velocity relative to the ice. I've tried to use:

    Pt = Pt'
    0 = Pt'
    0 = m1v1' + m2v2'
    -m1v1' = m2v2'
    -(100)v1' = (20)(8)
    v1' = -1.6 m/s

    But it still doesnt get me the right answer. I guess i'm just not very good with understanding the concept of relative velocity.
     
  10. Dec 3, 2006 #9

    OlderDan

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    0 = m1v1' + m2v2'
    v1' - v2' = 10m/s <== this is what relative velocity means
    -(100kg)v1' = (20kg)(v1' - 10m/s)

    In the more general case, velocity is a vector. Suppose you had a vehicle moving North at 30m/s and another moving East at 40m/s. What would be the velocity of the northbound vehicle relative to the eastbound vehicle? The relative velocity would be the rate at which they are separating in the direction from the eastbound to the northbound. The distance between them is

    d(t) = sqrt[(30m/s*t)² + (40m/s*t)²] = 50m/s*t

    The relative velocity is 50m/s. The direction is somewhat North of West. Representing the positions as vectors you have the relative position

    r(t) = 30m/s*t j - 40m/s*t i

    with relative velocity

    v = 30m/s j - 40m/s i
     
  11. Dec 3, 2006 #10
    Wow, this is starting to make sense now. So going by your equation:

    -(100kg)v1' = (20kg)(v1' - 10m/s)
    then v1' = 1.667 m/s Which is the speed of Jean relative to the ice after he throws the box.

    And v2' would be:
    0 = m1v1' + m2v2'
    0 = (100)(1.667) + (20)v2'
    v2' = -8.335 m/s Which is the speed of the box relative to the ice.

    So the speed of Lucien would be:

    m1v1 + m2v2 = v'(m1 + m2)
    (20)(-8.335) + (100)(0) = v'(20 + 100)
    v' = -1.39 m/s

    And their relative separation speed would be:

    v = v' - (v1')
    v = (1.67) - (-1.39)
    v = 3.06 m/s

    So the correct answer is b) 3.1 m/s

    Right?
     
    Last edited: Dec 3, 2006
  12. Dec 4, 2006 #11
    Well I've handed it in now and I'd like to thank you OlderDan for all your help. It was greatly appreciated.
     
  13. Feb 15, 2011 #12
    Man this is simple phsics just ask me any questiones and i will answer them and also u can incude impulse; acceleration, velocity, and time problems; force problems; moment problems; and optics problems;
     
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